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liq [111]
3 years ago
14

Cody’s car accelerates from 0m/s to 45 m/s northward in 15 seconds. What is the acceleration of the car

Engineering
1 answer:
sasho [114]3 years ago
7 0

Answer:

3 m/s²

Explanation:

Acceleration is calculated as :

a= Δv/ t

where ;

Δv = change in velocity

Δv = 45 - 0 = 45  m/s

t= 15 s

a= 45 /15

a= 3 m/s²

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A right triangle has a base of 12 inches and a height of 30 inches, what is the centroid of the triangle?​
aliina [53]

Answer:

the correct answer is 42

4 0
2 years ago
The pads are 200mm long, 150 mm wide and thickness equal to 12mm. 1- Determine the average shear strain in the rubber if the for
lord [1]

Answer:

a) 0.3

b) 3.6 mm

Explanation:

Given

Length of the pads, l = 200 mm = 0.2 m

Width of the pads, b = 150 mm = 0.15 m

Thickness of the pads, t = 12 mm = 0.012 m

Force on the rubber, P = 15 kN

Shear modulus on the rubber, G = 830 GPa

The average shear strain can be gotten by

τ(average) = (P/2) / bl

τ(average) = (15/2) / (0.15 * 0.2)

τ(average) = 7.5 / 0.03

τ(average) = 250 kPa

γ(average) = τ(average) / G

γ(average) = 250 kPa / 830 kPa

γ(average) = 0.3

horizontal displacement,

δ = γ(average) * t

δ = 0.3 * 12

δ = 3.6 mm

5 0
3 years ago
What do you own that might not be manufactured?
horrorfan [7]

Answer:

A pet

Explanation:

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8 0
3 years ago
Maximum iorsional shear siress.? Select one: a)- occurs at the center of a shaft. b)- occurs at the outer surface of a shaft c)-
pickupchik [31]

Answer:

b). Occurs at the outer surface of the shaft

Explanation:

We know from shear stress and torque relationship, we know that

\frac{T}{J}= \frac{\tau }{r}

where, T = torque

            J = polar moment of inertia of shaft

            τ = torsional shear stress

             r = raduis of the shaft

Therefore from the above relation we see that

            \tau = \frac{T.r}{J}

Thus torsional shear stress, τ is directly proportional to the radius,r of the shaft.

When r= 0, then τ = 0

and when r = R , τ is maximum

Thus, torsional shear stress is maximum at the outer surface of the shaft.

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3 years ago
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leonid [27]

Answer=

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Explanation:

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3 years ago
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